Is this the correct graph of the equation y =
52?
(1 point)
No, it is not the correct
O graph because the slope m =
rise =
run
5°
Yes, it is the correct graph because the ordered pair
• (0, 0) satisfies the equation and the
slope m = rise = 1
run
5°
Yes, it is the correct graph
O because the slope = 5.
No, it is not the correct graph. The equation y = 52 represents a horizontal line that passes through the point (0, 52). The graph should be a straight line parallel to the x-axis, and not a line with a slope of 5.
Based on the information given, it is not possible to determine whether the graph shown is the correct graph of the equation y = 52. The details provided do not give enough information about the slope or any specific points on the graph.
In order to determine if the given graph is correct for the equation y = 52, we need to analyze the characteristics of the equation and its graph.
The equation y = 52 represents a horizontal line parallel to the x-axis, where the y-coordinate is always 52, regardless of the value of x. This means that every point on the graph will have a y-coordinate of 52.
Looking at the given options:
- Option 1 states that the slope (m) is 5°. However, since the equation represents a horizontal line, the slope is 0, not 5°. Therefore, option 1 is incorrect.
- Option 2 mentions that the ordered pair (0,0) satisfies the equation and the slope (m) equals the ratio rise/run, which is 1/5. However, since the equation is y = 52, the ordered pair (0,0) does not satisfy the equation. Therefore, option 2 is incorrect.
- Option 3 simply states that the slope is 5. However, as mentioned earlier, the slope of the equation y = 52 is 0, not 5. Thus, option 3 is also incorrect.
Therefore, none of the options provided state the correct reason for the graph being correct or incorrect. The correct answer should be: No, it is not the correct graph because the equation y = 52 represents a horizontal line with a constant y-coordinate of 52, whereas the given graph appears to have a non-constant y-coordinate.